1. Kinetics Part 1
22 Nov. 2010
Objective: SWBAT write rate expressions and calculate reaction rates for chemical reactions.
Do now: Describe one very slow reaction that you’ve seen, and one very fast reaction.

2. Agenda Do now Kinetics notes Reaction Rates Demonstrations
Kinetics Part 1
Agenda
Do now
Kinetics notes
Reaction Rates Demonstrations
Rate constant and reaction rates problems.
Homework: p. 602 #2, 3, 5, 7, 12, 13, 15, 17, 19

3. How can we predict whether or not a reaction will take place?
Kinetics Part 1
How can we predict whether or not a reaction will take place?
Thermodynamics
Once started, how fast does the reaction proceed?
Chemical kinetics: this unit!
How far will the reaction go before it stops?
Equilibrium: next unit

4. Kinetics Part 1
Chemical Kinetics
The area of chemistry concerned with the speeds, or rates, at which a chemical reaction occurs.
reaction rate: the change in the concentration of a reactant or product with time (M/s)
Why do reactions have such very different rates?
Steps in vision: to 10-6 seconds!
Graphite to diamonds: millions of years!
In chemical industry, often more important to maximize the speed of a reaction, not necessarily yield.

5. Kinetics Part 1
Chemical Kinetics
Reaction rate is the change in the concentration of a reactant or a product with time (M/s).
A B
rate = -
D[A] Dt
D[A] = change in concentration of A over
time period Dt
rate =
D[B] Dt
D[B] = change in concentration of B over
time period Dt
Because [A] decreases with time, D[A] is negative.

6. A B
rate = -
D[A] Dt
rate =
D[B] Dt

7. Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
red-brown
Br2 (aq) + HCOOH (aq) Br- (aq) + 2H+ (aq) + CO2 (g)
time
393 nm
light
Detector
t1< t2 < t3
D[Br2] a D Absorption

8. Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)
slope of
tangent
slope of
tangent
slope of
tangent
average rate = -
D[Br2] Dt
= -
[Br2]final – [Br2]initial
tfinal - tinitial
instantaneous rate = rate for specific instance in time

9. rate a [Br2]
rate = k [Br2]
k =
rate
[Br2]
= rate constant
= 3.50 x 10-3 s-1

10. measure DP over time
2H2O2 (aq) H2O (l) + O2 (g)
PV = nRT
P = RT = [O2]RT n V
[O2] = P RT 1
rate =
D[O2] Dt RT 1 DP Dt =

12. Reaction Rates and Stoichiometry
2A B
Two moles of A disappear for each mole of B that is formed.
rate = -
D[A] Dt 1 2
rate =
D[B] Dt
aA + bB cC + dD
rate = -
D[A] Dt 1 a
= -
D[B] Dt 1 b =
D[C] Dt 1 c =
D[D] Dt 1 d

13. Example Write the rate expression for the following reaction:
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)

14. Write the rate expression for the following reaction:
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
rate = -
D[CH4] Dt
= -
D[O2] Dt 1 2 =
D[CO2] Dt =
D[H2O] Dt 1 2

15. Practice Problems
Write the rate expressions for the following reactions in terms of the disappearance of the reactants and appearance of products.
I-(aq) + OCl-(aq)  Cl-(aq) + OI-(aq)
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)

16. Using Rate Expressions
Consider the reaction:
4NO2(g) + O2(g)  2N2O5(g)
Suppose that, at a particular moment during the reaction, molecular oxygen is reacting at the rate of M/s.
At what rate is N2O5 being formed?
At what rate is NO2 reacting?

17. Consider the reaction:
4PH3(g)  P4(g) + 6H2(g)
Suppose that, at a particular moment during the reaction, molecular hydrogen is being formed at the rate of M/s.
At what rate is P4 being formed?
At what rate is PH3 reacting?

18. The Rate Law
The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers.
aA + bB cC + dD
Rate = k [A]x[B]y
Reaction is xth order in A
Reaction is yth order in B
Reaction is (x +y)th order overall
x and y are determined experimentally!

19. F2 (g) + 2ClO2 (g) 2FClO2 (g) rate = k [F2]x[ClO2]y
Double [F2] with [ClO2] constant
Rate doubles
x = 1
Quadruple [ClO2] with [F2] constant
Rate quadruples
y = 1
rate = k [F2][ClO2]

20. Rate Laws Rate laws are always determined experimentally.
Reaction order is always defined in terms of reactant (not product) concentrations.
The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.
F2 (g) + 2ClO2 (g) FClO2 (g) 1
rate = k [F2][ClO2]

21. S2O82- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq)
Determine the rate law and calculate the rate constant for the following reaction from the following data:
S2O82- (aq) + 3I- (aq) SO42- (aq) + I3- (aq)
Experiment
[S2O82-]
[I-]
Initial Rate (M/s) 1
0.08
0.034
2.2 x 10-4 2
0.017
1.1 x 10-4 3
0.16

22. S2O82- (aq) + 3I- (aq) 2SO42- (aq) + I3- (aq)
Determine the rate law and calculate the rate constant for the following reaction from the following data:
S2O82- (aq) + 3I- (aq) SO42- (aq) + I3- (aq)
Experiment
[S2O82-]
[I-]
Initial Rate (M/s) 1
0.08
0.034
2.2 x 10-4 2
0.017
1.1 x 10-4 3
0.16
rate = k [S2O82-]x[I-]y
y = 1
x = 1
rate = k [S2O82-][I-]
Double [I-], rate doubles (experiment 1 & 2)
Double [S2O82-], rate doubles (experiment 2 & 3)
k =
rate
[S2O82-][I-] =
2.2 x 10-4 M/s
(0.08 M)(0.034 M)
= 0.08/M•s

23. Practice Problems
The reaction of nitric oxide with hydrogen at 1280oC:
2NO(g) + 2H2(g)  N2(g) + 2H2O(g)
From the following data collected at this temperature, determine (a) the rate law, (b) the rate constant and (c) the rate of the reaction when [NO] = 12.0x10-3 M and [H2] = 6.0x10-3 M
Experiment
[NO] M
[H2] M
Initial Rate (M/s) 1
5.0x10-3
2.0x10-3
1.3x10-5 2
10.0x10-3
5.0x10-5 3
4.0x10-3
10.0x10-5

24. 23 Nov. 2010
Objective: SWBAT write rate expressions and calculate reaction rates for chemical reactions.
Do now: Calculate the rate constant (k) from the data below:

25. Agenda Do now Practice Problems
Homework: Revisit last night’s assignment
#19 hint: Graph lnP vs. time. If linear, it is 1st order. slope = -k
Excel works great.

26. 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
Kinetics Part 1
Write the reaction rate expressions for the following in terms of the disappearance of the reactants and the appearance of products:
2H2(g) + O2(g)  2H2O(g)
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)

27. At what rate is ammonia being formed?
Kinetics Part 1
Consider the reaction
N2(g) + 3H2(g)  2NH3(g)
Suppose that at a particular moment during the reaction molecular hydrogen is reacting at a rate of M/s.
At what rate is ammonia being formed?
At what rate is molecular nitrogen reacting?

28. Kinetics Part 1
Calculate the rate of the reaction at the time when [F2] = M and [ClO2] = M.
F2(g) + 2ClO2(g)  2FClO2(g)
[F2] (M)
[ClO2] (M)
Initial Rate (M/s)
0.10
0.010
1.2x10-3
0.040
4.8x10-3
0.20
2.4x10-3

29. Consider the reaction X + Y  Z
Kinetics Part 1
Consider the reaction X + Y  Z
From the following data, obtained at 360 K,
determine the order of the reaction
determine the initial rate of disappearance of X when the concentration of X is 0.30 M and that of Y is 0.40 M
Initial Rate of Disappearance of X (M/s)
[X] (M)
[Y] (M)
0.053
0.10
0.50
0.127
0.20
0.30
1.02
0.40
0.60
0.254
0.509

30. Consider the reaction A B.
Kinetics Part 1
Consider the reaction A B.
The rate of the reaction is 1.6x10-2 M/s when the concentration of A is 0.35 M. Calculate the rate constant if the reaction is
first order in A
second order in A

31. 24 Nov. 2010
Objective: SWBAT determine reaction order graphically and relate concentration of 1st order reactions to time.
Do now: What is the overall order of reaction for the data shown below?
What is the rate constant?
Run #
Initial [A] ([A]0)
Initial [B] ([B]0)
Initial Rate (v0) 1
1.00 M
1.25 x 10-2 M/s 2
2.00 M
2.5 x 10-2 M/s 3

32. Agenda Do now Homework solutions First order reactions
Time calculations for first-order reactions.
Homework: p. 603 #19, 21, 22, 23, 25-29

33. The rate laws can be used to determine the concentrations of any reactants at any time during the course of a reaction.

34. 29 Nov. 2010 Take Out Homework p. 603 #19, 21, 22, 23, 25- 29
Objective: SWBAT compare 1st order, 2nd order, and zero order reactions, and describe how temperature and activation energy effect the rate constant.
Do now: Calculate the half life of the reaction F2(g) + 2ClO2(g)  2FClO2(g), with rate data shown below:
[F2] (M)
[ClO2] (M)
Initial Rate (M/s)
0.10
0.010
1.2x10-3
0.040
4.8x10-3
0.20
2.4x10-3

35. Agenda Homework solutions Review 1st order reactions
Second and zero order reactions
Activation Energy
Problem Set
Quiz Weds.
p. 31, 32, 35, 37, 39, 42
+ Problem set

36. First-Order Reactions
rate = -
D[A] Dt
A product
rate = k [A]
D[A] Dt
= k [A] -
rate
[A]
M/s M =
k =
= 1/s or s-1
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
[A] = [A]0e−kt
ln[A] = ln[A]0 - kt

37. Graphical Determination of k
2N2O NO2 (g) + O2 (g)

38. The reaction 2A B is first order in A with a rate constant of 2
The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ?

39. The reaction 2A B is first order in A with a rate constant of 2
The reaction 2A B is first order in A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ?
[A]0 = 0.88 M
ln[A] = ln[A]0 - kt
[A] = 0.14 M
kt = ln[A]0 – ln[A] ln
[A]0
[A] k = ln
0.88 M
0.14 M
2.8 x 10-2 s-1 =
ln[A]0 – ln[A] k
t =
= 66 s

40. The conversion of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7x10-4 s-1 at 500oC.
If the initial concentration of cyclopropane was M, what is the concentration after 8.8 minutes?
How long, in minutes, will it take for the concentration of cyclopropane to decrease from 0.25 M to 0.15 M?
How long, in minutes, will it take to convert 74% of the starting material?

41. Practice Problem
The reaction 2A → B is first order in A with a rate constant of 2.8×10-2s-1 at 80oC. How long, in seconds, will it take for A to decrease from 0.88 M to 0.14 M?

42. The rate of decomposition of azomethane (C2H6N2) is studied by monitoring partial pressure of the reactant as a function of time: CH3-N=N-CH3(g) → N2(g) + C2H6(g) The data obtained at 300oC are shown here: Are these values consistent with first-order kinetics? If so, determine the rate constant.
Time (s)
Partial Pressure of Azomethane (mmHg)
284
100
220
150
193
200
170
250
300
132

43. The following gas-phase reaction was studied at 290oC by observing the change in pressure as a function of time in a constant-volume vessel:
ClCO2CCl3(g)  2COCl2(g)
Determine the order of the reaction and the rate constant based on the following data, where P is the total pressure
Time (s)
P (mmHg)
400
2,000
316
4,000
248
6,000
196
8,000
155
10,000
122

44. Ethyl iodide (C2H5I) decomposes at a certain temperature in the gas phase as follows: C2H5I(g) → C2H4(g) + HI(g) From the following data, determine the order of the reaction and the rate constant:
Time (min)
[C2H5I] (M)
0.36 15
0.30 30
0.25 48
0.19 75
0.13

45. First-Order Reactions
The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.
t½ = t when [A] = [A]0/2 ln
[A]0
[A]0/2 k = t½
ln 2 k =
0.693 k =
What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1?
How do you know decomposition is first order?

46. First-Order Reactions
The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.
t½ = t when [A] = [A]0/2 ln
[A]0
[A]0/2 k = t½
ln 2 k =
0.693 k =
What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1? t½
ln 2 k =
0.693
5.7 x 10-4 s-1 =
= 1200 s = 20 minutes
How do you know decomposition is first order?
units of k (s-1)

47. First-order reaction
A product
# of
half-lives
[A] = [A]0/n 1 2 2 4 3 8 4 16

48. The decomposition of ethane (C2H6) to methyl radicals is a first-order reaction with a rate constant of 5.36x10-4 s-1 at 700oC:
C2H6(g)  2CH3(g)
Calculate the half-life of the reaction in minutes.

49. Calculate the half-life of the decomposition of N2O5:
2N2O5  4NO2(g) + O2(g)
t (s)
[N2O5] (M)
ln [N2O5]
0.91
-0.094
300
0.75
-0.29
600
0.64
-0.45
1200
0.44
-0.82
3000
0.16
-1.83

50. Second-Order Reactions
rate = -
D[A] Dt
A product
rate = k [A]2
rate
[A]2
M/s M2 =
D[A] Dt
= k [A]2 -
k =
= 1/M•s 1
[A] =
[A]0
+ kt
[A] is the concentration of A at any time t
[A]0 is the concentration of A at time t=0
t½ = t when [A] = [A]0/2
t½ = 1
k[A]0

51. Iodine atoms combine to form molecular iodine in the gas phase:
I(g) + I(g)  I2(g)
This reaction follows second-order kinetics and has the high rate constant 7.0x109/M·s at 23oC.
If the initial concentration of I was M, calculate the concentration after 2.0 minutes.
Calculate the half-life of the reaction if the initial concentration of I is 0.60 M and if it is 0.42 M.

52. Calculate the half-life of the reaction.
Kinetics Part 1
The reaction 2A → B is second order with a rate constant of 51/M·min at 24oC.
Starting with [A]o = M, how long will it take for [A]t = 3.7x10-3 M?
Calculate the half-life of the reaction.

53. Zero-Order Reactions rate = - D[A] Dt A product rate = k [A]0 = k rate
= M/s
[A] is the concentration of A at any time t
[A] = [A]0 - kt
[A]0 is the concentration of A at time t = 0
t½ = t when [A] = [A]0/2
t½ =
[A]0 2k

54. Summary of the Kinetics of Zero-Order, First-Order
and Second-Order Reactions
Order
Rate Law
Concentration-Time Equation
Half-Life
t½ =
[A]0 2k
rate = k
[A] = [A]0 - kt t½
ln 2 k = 1
rate = k [A]
ln[A] = ln[A]0 - kt 1
[A] =
[A]0
+ kt
t½ = 1
k[A]0 2
rate = k [A]2

55. Activation Energy and Temperature Dependence of Rate Constants
Reaction rates increase with increasing temperature
Ex: Hard boiling an egg
Ex: Storing food
How do reactions get started in the first place?

56. Collision Theory
Chemical reactions occur as a result of collisions between reacting molecules.
reaction rate depends on concentration
But, the relationship is more complicated than you might expect!
Not all collisions result in reaction (p )

57. In order to react, colliding molecules must have a total KE ≥ activation energy (Ea)
Ea: minimum amount of energy required to initiate a chemical reaction
activated complex (transition state): a temporary species formed by the reactant molecules as a result of the collision before they form the product.

58. A + B AB C + D +
Exothermic Reaction
Endothermic Reaction
The activation energy (Ea ) is the minimum amount of energy required to initiate a chemical reaction.
=a barrier that prevents less energetic molecules from reacting

59. Temperature Dependence of the Rate Constant
(Arrhenius equation)
Ea is the activation energy (J/mol)
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature
A is the frequency factor (constant
across wide range of temps.)
Alternate format (y=mx+b):
ln k = - Ea R 1 T
+ lnA

60. Alternate Form of the Arrhenius Equation
At two temperatures, T1 and T2 or

61. The rate constants for the decomposition of acetaldehyde: CH3CHO(g) → CH4(g) + CO(g) were measured at five different temperatures. The data are shown below. Plot lnk versus 1/T, and determine the activation energy (in kJ/mol) for the reaction. (Note: the reaction is order in CH3CHO, so k has the units of ) k
T (K)
0.011
700
0.035
730
0.105
760
0.343
790
0.789
810

62. slope = -2.19x104 K
slope =

63. The second order rate constant for the decomposition of nitrous oxide (N2O) into nitrogen molecule and oxygen atom has been measured at different temperatures. Determine graphically the activation energy for the reaction. k
T (oC)
1.87x10-3
600
0.0113
650
0.0569
700
0.244
750

64. 30 Nov. 2010 Take Out Homework p. 605# 31, 32, 35, 37, 39
Objective: SWBAT use the Arrhenius equation to solve for rate constants and temperatures, and solve practice problems on kinetics.
Do now: Match
Order
Rate Law
Conc-Time Eq.
Half Life Eq. 2
rate = k[A]
[A]=[A]0-kt
t1/2=1/k[A]o 1
rate = k[A]2
1/[A]=1/[A]0 + kt
t1/2=ln2/k
rate = k
ln[A]=ln[A]0 –kt
t1/2=[A]0/2k

65. Agenda Homework solutions Using the Arrhenius equation part 2
Molecular orientation
Problem Set work time
Homework: Complete problem set and
p. 605 #40, 42
Quiz tomorrow

66. The rate constant of a first order reaction is 3. 46x10-2 /s at 298 K
The rate constant of a first order reaction is 3.46x10-2 /s at 298 K. What is the rate constant at 350 K if the activation energy for the reaction is 50.2 kJ/mol?

67. The first order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochloric acid (HCl) is 3.32x10-10/s at 25oC. Calculate the rate constant at 40oC if the activation energy is 116 kJ/mol.

68. Homework
Quiz Weds.
p. 604 #40, 42
+ Problem set #2, 18, 20, 25, 29, 32, 37, 42, 55, 57: Weds.

69. Problem Set Handout #2, 18, 20, 25, 29, 32, 37, 42, 55, 57: Weds.
Quiz Weds.

70. 2 Dec. 2010
Objective: SWBAT describe a reaction in terms of elementary steps and determine the rate law.
Do now: Which graph represents an exothermic reaction? Which is endothermic? Why?

71. Agenda Do now Elementary steps in reactions Catalysts
Homework: p. 605 #45, 47, 48, 52, 54, 55, 56, 58, 62: Mon.

72. For simple reactions (between atoms, for example) the frequency factor (A) is proportional to the frequency of collision between the reacting species.
“Orientation factor” is also important.

73. Importance of Molecular Orientation
effective collision
ineffective collision

74. A balanced chemical equation doesn’t tell us much about how the reaction actually takes place.
It may represent the sum of elementary steps
Reaction mechanism: the sequence of elementary steps that leads to product formation.

75. N2O2 is detected during the reaction!
Reaction Mechanisms
The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions.
The sequence of elementary steps that leads to product formation is the reaction mechanism.
2NO (g) + O2 (g) NO2 (g)
N2O2 is detected during the reaction!
Elementary step:
NO + NO N2O2
Overall reaction:
2NO + O NO2 +
Elementary step:
N2O2 + O NO2

76. 2NO (g) + O2 (g) NO2 (g)
Mechanism:

77. Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation.
An intermediate is always formed in an early elementary step and consumed in a later elementary step.
Elementary step:
NO + NO N2O2
N2O2 + O NO2
Overall reaction:
2NO + O NO2 +
The molecularity of a reaction is the number of molecules reacting in an elementary step.
Unimolecular reaction – elementary step with 1 molecule
Bimolecular reaction – elementary step with 2 molecules
Termolecular reaction – elementary step with 3 molecules

78. Rate Laws and Elementary Steps
Unimolecular reaction
A products
rate = k [A]
Bimolecular reaction
A + B products
rate = k [A][B]
Bimolecular reaction
A + A products
rate = k [A]2
Writing plausible reaction mechanisms:
The sum of the elementary steps must give the overall balanced equation for the reaction.
The rate-determining step should predict the same rate law that is determined experimentally.
The rate-determining step is the slowest step in the sequence of steps leading to product formation.

79. Sequence of Steps in Studying a Reaction Mechanism

80. The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps:
Step 1:
NO2 + NO NO + NO3
Step 2:
NO3 + CO NO2 + CO2
What is the equation for the overall reaction?
What is the intermediate?
What can you say about the relative rates of steps 1 and 2?

81. What is the equation for the overall reaction?
The experimental rate law for the reaction between NO2 and CO to produce NO and CO2 is rate = k[NO2]2. The reaction is believed to occur via two steps:
Step 1:
NO2 + NO NO + NO3
Step 2:
NO3 + CO NO2 + CO2
What is the equation for the overall reaction?
NO2+ CO NO + CO2
What is the intermediate?
NO3
What can you say about the relative rates of steps 1 and 2?
rate = k[NO2]2 is the rate law for step 1 so
step 1 must be slower than step 2

82. rate determining step: the slowest step in the sequence of steps leading to product formation.

83. Decomposition of Hydrogen Peroxide
2H2O2(aq)  2H2O(l) + O2(g) Can be catalyzed using iodide ions (I-) rate = k[H2O2][I-] Why?! Determined experimentally. Step 1: H2O2 + I-  H2O + IO- Step 2: H2O2 + IO-  H2O + O2 + I-

84. 7 Dec. 2010
Objective: SWBAT describe catalysis and how ammonia is synthesized through the Haber process. Do now: The rate for the reaction 2H2O2(aq)  2H2O(l) + O2(g) is rate = k[H2O2][I-] Step 1: H2O2 + I-  H2O + IO- Step 2: H2O2 + IO-  H2O + O2 + I- Which is the rate determining step? Why?

85. Agenda Do now Homework solutions Catalysts and catalyzing mechanisms
AP Kinetics problem set
Homework: p. 605 #55, 56, 58
Bring your book on Thursday!

86. Example
The gas-phase decomposition of nitrous oxide (N2O) is believed to occur via two elementary steps:
Step 1: N2O  N2 + O
Step 2 N2O + O  N2 + O2
Experimentally the rate law is found to be
rate = k[N2O].
Write the equation for the overall reaction.
Identify the intermediates.
What can you say about the relative rates of steps 1 and 2?

87. For the decomposition for H2O2, the reaction rate depends on the concentration of I- ions, even though I- doesn’t appear in the overall equation.
I- is a catalyst for the reaction.

88. A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. Ea k
Catalyzed
Uncatalyzed
ratecatalyzed > rateuncatalyzed
Ea < Ea ′

89. Catalysts forms an alternative reaction pathway
for example, it might form an intermediate with the reactant.
Ex: 2KClO3(s)  2KCl(s) + 3O2(g)
Very slow, until you add MnO2, a catalyst. The MnO2 can be recovered at the end of the reaction!

90. In heterogeneous catalysis, the reactants and the catalysts are in different phases (usually, catalyst is a solid, reactants are gases or liquids).
Haber synthesis of ammonia
Ostwald process for the production of nitric acid
Catalytic converters
In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid.
Acid catalysis
Base catalysis

91. Extremely slow at room temperature. Must be fast and high yield!
Haber Process
Synthesis of Ammonia
Extremely slow at room temperature. Must be fast and high yield!
Process occurs on the surface of the Fe/Al2O3/K2O catalyst, which weakens the covalent N-N and H-H bonds.
Fe/Al2O3/K2O
catalyst
N2 (g) + 3H2 (g) NH3 (g)

92. Ostwald Process 4NH3 (g) + 5O2 (g) 4NO (g) + 6H2O (g)
Pt catalyst
4NH3 (g) + 5O2 (g) NO (g) + 6H2O (g)
2NO (g) + O2 (g) NO2 (g)
2NO2 (g) + H2O (l) HNO2 (aq) + HNO3 (aq)
Pt-Rh catalysts used
in Ostwald process

93. Catalytic Converters CO + Unburned Hydrocarbons + O2 CO2 + H2O
2NO + 2NO2
2N2 + 3O2

94. Enzyme Catalysis
biological catalysts

95. Binding of Glucose to Hexokinase

96. p. 605 #45, 47, 48, 52, 54, 55, 56, 58, 62